A pplied Thermodynamics

Applied Thermodynamics

3. GAS TURBINES AND JET PROPULSION Introduction:

Gas turbines are prime movers producing mechanical power from the heat generated by the combustion of fuels.

They are used in aircraft, some automobile units, industrial installations and small – sized electrical power generating units.

A schematic diagram of a simple gas turbine power plant is shown below.

This is the open cycle gas turbine plant.

Working: Air from atmosphere is

compressed adiabatically (idealized) in a compressor (usually rotary) i.e., Process 1–2.

This compressed air enters the combustion chamber, where fuel is injected and undergoes combustion at constant pressure in process 2–3.

The hot products of combustion expand in the turbine to the ambient pressure in process 3–4 and the used up exhaust gases are let out into the surroundings.

The compressor is usually coupled to the turbine, so that the work input required by the compressor comes from the turbine.

The turbine produces more work than what is required by the compressor, so that there is net work output available from the turbine.Since the products of combustion cannot be re–used, real gas turbines work essentially in open cycles. The p–v and T–s diagrams of such a plant are shown above.

The compressor is usually coupled to the turbine, so that the work input required by the compressor comes from the turbine.

The turbine produces more work than what is required by the compressor, so that there is net work output available from the turbine.Since the products of combustion cannot be re–used, real gas turbines work essentially in open cycles. The p–v and T–s diagrams of such a plant are shown above.

Brayton Cycle: This is the air–

standard cycle for the gas turbine plant.

It consists of two reversible adiabatic processes and two reversible isobars (constant pressure processes).

The p–v and T–s diagrams of a Brayton Cycle are as shown.

Process 1 - 2: Reversible adiabatic compression.2 – 3: Reversible constant pressure heat addition.3 – 4: Reversible adiabatic expansion.4 – 1: Reversible constant pressure heat rejection.

A schematic flow diagram of this somewhat hypothetical gas turbine plant is shown below.

Though this plant works on a closed cycle, each of the four devices in the plant is a steady–flow device, in the sense that there is a continuous flow of the working fluid (air) through each device.

Hence, the steady–flow energy equation is the basis for analysis, and can be applied to each of the four processes.

Neglecting changes in kinetic and potential energies, the steady flow energy equation takes the from

Q – W = ∆h = Cp.∆T (Since air is assumed to be an ideal gas)

Process 1 – 2 is reversible adiabatic, hence Q1-2 = 0 W1-2 = - Cp.∆T = - Cp (T2 – T1): - ve, work input Work of compression Wc = |W1 – 2| = Cp (T2-T1)  

Process 2–3 is a constant pressure process

Heat added,

 Process 3-4 is again reversible

adiabatic,

+ve work output.

Process 4-1 is also a constant pressure process

: -ve, i.e., heat is rejected  Heat rejected, Q2 = |Q4-1| = Cp (T4-T1)

Therefore the cycle efficiency,

Therefore the cycle efficiency,

For isentropic process 1-2,

& for process 3-4,

Since p3 = p2 and p4 = p1

Compression ratio,

Therefore,

Pressure ratio,

Thus it can be seen that, for the same compression ratio,

A closed cycle turbine plant is used in a gas–cooled nuclear reactor plant, where the source is a high temperature gas cooled reactor supplying heat from nuclear fission directly to the working fluid (gas/air).

 Comparison between Brayton Cycle and Otto cycle:- For the same

compression ratio, and nearly same net work output (represented by the area inside the p–v diagram), the Brayton cycle handles larger range of volume and smaller range of pressure than does the Otto cycle.

A Brayton cycle is not suitable as the basis for

the working of reciprocating type of devices (Piston–Cylinder arrangements).

A reciprocating engine cannot efficiently handle a large volume flow of low pressure gas.

The engine (Cylinder) size becomes very large and friction losses become excessive.

Otto cycle therefore is more suitable in reciprocating engines.

However, a Brayton cycle is more suitable than an Otto cycle, as a basis for a turbine plant.

An I.C. engine is exposed to the highest temperature only intermittently (for short way during each cycle), so that there is time enough for it to cool.

On the other hand, a gas turbine, being a steady flow device, is continuously exposed to the highest temperature.

Metallurgical considerations, therefore limit the maximum temperature that can be used.

Moreover, in steady flow machines, it is easier to transfer heat at constant pressure than at constant volume.

Besides, turbines can be efficiently handle large volume of gas flow.

In view of all these, the Brayton cycle more suitable as the basis for the working of gas turbine plants.

Effect of irreversibility’s in turbine/compressor: In the ideal Brayton cycle,

compression and expansion of air are assumed to be reversible and adiabatic.

In reality, however, irreversibility’s do exist in the machine operations, even though they may be adiabatic.

Hence the compression and expansion processes are not really constant entropy processes.

Entropy tends to be increase (as per the principle of increase of entropy).

Effect of irreversibility’s in turbine/compressor: The T–s diagram of a Brayton

cycle subject to irreversibility’s will be as shown.

Irreversibility’s result in a reduction in turbine output by (h4-h4S) and in an increase in the compressor input by (h2 – h2S).

Hence the output reduces by the amount (h4–h4S )+ (h2–h2s).

Though heat input is also reduced by (h2-h2s), the cycle efficiency is less than that of an ideal cycle. The extent of losses due to irreversibility’s can be expressed in terms of the turbine and compressor efficiencies.

Turbine efficiency,

Compressor efficiency,

Methods of improving the efficiency of Brayton cycle:

  Use of regeneration: The efficiency of the Brayton cycle can be increased by

utilizing part of the energy of exhaust air from the turbine to preheat the air leaving the compressor, in a heat exchanger called regenerator.

This reduces the amount of heat supplied Q1 from an external source, and also the amount of heat rejected Q2 to an external sink, by an equal amount.

Since Wnet = Q1 - Q2 and both Q1 and Q2 reduce by equal amounts, there will be no change in the work output of the cycle.

Heat added Q1 = h3 –h2’ = Cp (T3 – T2’) Heat rejected Q2 = h4’ – h1 = Cp (T4’ – T1) Turbine output WT = h3 – h4 = Cp (T3 – T4) Compressor input WC = h2 – h1 = Cp (T2 – T1) Regeneration can be used only if the temperature of air

leaving the turbine at 4 is greater than that of air leaving the compressor at 2.

In the regenerator, heat is transferred from air leaving the turbine to air leaving the compressor, thereby raising the temperature of the latter.

The maximum temperature to which compressed air at 2 can be heated is equal to the temperature of turbine exhaust at 4.

This, however, is possible only in an ideal regenerator.

In reality, T2’<T4. The ratio of the

actual temperature rise of compressed air to the maximum possible rise is called effectiveness of the regenerator.

With a regenerator, since Wnet remains unchanged, but Q1 reduces, efficiency

η = Wnet/Q1 increases. This is also evident from the

fact that the mean temperature of heat addition increases and the mean temperature of heat rejection reduces with the use of the regenerator, and efficiency is also given by

With regenerator,

In the regenerator,Heat lost by hot air = Heat gained by cold air

i.e.,

With an ideal regenerator,

therefore,

For a fixed ratio , the cycle efficiency decreases

with increasing pressure ratio.

In practice, a regenerator is expensive, heavy

and bulky and causes pressure losses, which may

even decrease the cycle efficiency, instead of

increasing it.

2.Multistage compression with inter cooling:

In this arrangement, compression of air is carried out in two or more stages with cooling of the air in between the stages.

The cooling takes place in a heat exchanger using some external cooling medium (water, air etc).Shown above is a schematic flow diagram of a gas turbine plant with two-stage compression with inter cooling.  

1-2: first stage compression (isentropic)

2-3: inter cooling (heat rejection at

constant pressure)3-4: second stage

compression (isentropic)4-3: constant pressure heat

addition5-6: isentropic expansion6-1: constant pressure heat

rejection.

Air, after the first stage compression is cooled before it enters the second stage compressor.

If air is cooled to a temperature equal to the initial temperature (i.e., if T3=T1), inter cooling is said to be perfect.

In practice, usually T3 is greater than T1. Multistage compressor with inter cooling actually

decreases the cycle efficiency. This is because the average temperature of heat addition

Tadd is less for this cycle 1-2-3-4-5-6 as compared to the simple Brayton cycle 1-4’-5-6 with the initial state 1. (refer

fig). Average temperature of heat rejection Trej also reduces,

but only marginally.  

Hence efficiency is less for the modified cycle. However, if a regenerator is also used the heat added at lower temperature range (4 to 4’) comes from exhaust gases from the turbine.

So there may be an increase in efficiency (compared to a simple Brayton cycle) when multi–stage compression with inter cooling is used in conjunction with a regenerator.

For a gas turbine plant using 2–stage compression without a generator,

  Q1 = h5 - h4 = Cp(T5 - T4) WT = h5 - h6 = Cp(T5-T6) WC = (h2 - h1) + (h4 - h3) = Cp [(T2 - T1) + (T4 - T3)] 

WC = (h2 - h1) + (h4 - h3) = Cp [(T2 - T1) + (T4 - T3)]  Wnet = WT – WC

= Cp [(T5 - T6) – {(T 2- T1) + (T4 - T3)}]

3) Multi-Stage expansion with reheating:   

  Here expansion of working fluid (air) is carried out in 2 or more stages with heating (called reheating) in between stages.

The reheating is done in heat exchangers called Reheaters.

In an idealized cycle, the air is reheated, after each stage of expansion, to the temperature at the beginning of expansion.

The schematic flow diagram as well as T-s diagram for a gas turbine plant where in expansion takes place in two turbine stages, with reheating in between, are shown.

Multi-Stage expansion with reheating, by itself, does not lead to any improvement in cycle efficiency. In fact, it only reduces.

However, this modification together with regeneration may result in an increase in cycle efficiency.

It can be seen from the T-s diagram that the turbine exhaust temperature is much higher when multi stage expansion with reheating is used, as compared to a simple Brayton cycle.

This makes the use of a regenerator more effective and may lead to a higher efficiency.

Heat added Q1 = (h3 - h2) + (h5 - h4)   = Cp(T3 - T2) + Cp(T5 - T4)  Turbine output WT = (h3 - h4) + (h5 - h6) = Cp(T3 - T4) + Cp(T5 - T6) Compressor input WC = h2 - h1 = Cp(T2 - T1) 

Ideal Regenerative cycle with inter cooling and reheat:

Considerable improvement in efficiency is possible by incorporating all the three modifications simultaneously. Let us consider a regenerative gas turbine cycle with two stage compression and a single reheat.

The flow diagram and T-S diagram of such an arrangement is shown.Idealized Regenerative Brayton cycle with two stage compression with inter cooling and also two stage expansion with reheating – ideal regenerator, equal pressure ratios for stages, no irreversibilities, perfect inter cooling and reheating.

Heat added Q1 = Cp(T5 - T4’) + Cp(T7 - T6) Turbine output WT = Cp(T5 - T6) + Cp(T7 - T8) Compressor input WC = Cp(T2 - T1) + Cp(T4 - T3) 

If perfect inter cooling, no irreversibilities, equal pressure ratios for stages and ideal regenerator are assumed,

T1=T3, T2=T4=T8’, T5=T7 and T6=T8=T4’

Then, Q1 = Cp(T5 - T4’) + Cp(T7 – T6) = Cp (T5 - T6) + Cp(T5 - T6) = (T5 - T6)

Q2 = Cp(T8’ - T1) + Cp(T2 - T3) = Cp(T2 - T1) + Cp(T2 - T1) =2 Cp(T2 - T1)  

.

It can be seen from this expression that the efficiency decreases with increasing pressure ratio rp.

Effect of pressure Ratio rp on simple Brayton Cycle:-

  That means, the more the pressure ratio, the more will be the efficiency. Temperature T1 (=Tmin) is dependent on the temperature of surroundings.

Temperature T3 (=Tmax) is limited by metallurgical considerations and heat resistant characteristics of the turbine blade material.

For fixed values of Tmin and Tmax, the variation in net work output, heat added and efficiency with increasing pressure ratio rp can be explained with the help of a T-s diagram as shown.

For low pressure ratio, the net work output is small and the efficiency is also small (Cycle 1 – 2 – 3 - 4).

In the limit, as rp tends 1, efficiency tends to zero (net work output is zero, but heat added is not zero).

As the pressure ratio increases, the work output increases and so does the efficiency.

However, there is an upper limit for rp when the compression ends at Tmax.

As rp approaches this upper limit (rp)max, both net work output and heat added approach zero values.

However, it can be seen that the mean temperature heat addition Tadd approaches Tmax, while the mean temperature of heat rejection approaches Tmin, as rp

comes close to (rp)max.

Hence cycle efficiency, given by

approaches the Carnot efficiency i.e.,

rp - (rp)max When the compression ends at Tmax i.e., when state point 2 is at Tmax.

When rp=rpmax,

The variation of net work output Wnet with pressure ratio rp is shown below.

As rp increases from 1 to (rp)max, Wnet increases from zero, reaches a maximum at an optimum value of rp i.e., (rp)opt and with further increase in rp, it reduces and becomes zero when rp = rpmax

Pressure Ratio for maximum net work output:- Wnet= Cp[(T3 - T4) - (T2 - T1)]  T3 = Tmax & T1= Tmin

Condition for maximum Wnet is  

i.e.,

It can be seen that,

Maximum net work output

Corresponding to rp = (rp)opt i.e., when Wnet is maximum, cycle efficiency is

Open Cycle Gas Turbine Plants:

In practice, a gas turbine plant works on an open cycle.

Air from atmosphere is first compressed to a higher pressure in a rotary compressor, which is usually run by the turbine itself, before it enters the combustion chamber.

Fuel is injected into the combustion chamber where it undergoes combustion.

The heat released is absorbed by the products of combustion and the resulting high temperature; high pressure products expand in the turbine producing work output.

The used up combustion products (exhaust gases) are let out into the atmosphere.

In the ideal case, compression and expansion are assumed to be isentropic and combustion is assumed to take place at constant pressure.

The schematic flow diagram and p-v and T-s diagrams of an open cycle gas turbine plant are as shown.

Advantages and disadvantages of closed cycle over open cycle 

Advantages of closed cycle: 1.Higher thermal efficiency2.Reduced size3.No contamination4. Improved heat transmission5. Improved part load 6.Lesser fluid friction7.No loss of working medium8.Greater output and 9. Inexpensive fuel.

Disadvantages of closed cycle: 1.Complexity2.Large amount of cooling water is required. This

limits its use of stationary installation or marine use

3.Dependent system4.The wt of the system pre kW developed is high

comparatively, not economical for moving vehicles

5.Requires the use of a very large air heater.

Problems: 1. In a Gas turbine installation, the air is taken in at 1

bar and 150C and compressed to 4 bar. The isentropic of turbine and the compressor are 82% and 85% respectively. Determine (i) compression work, (ii) Turbine work, (iii) work ratio, (iv) Th. .

  What would be the improvement in the th. if a regenerator with 75% effectiveness is incorporated in the cycle. Assume the maximum cycle temperature to be 8250K.

 Solution: P1 = 1 bar T1 = 2880K P2 = 4 bar

T3 = 8250K C = 0.85 t = 0.82

Case1: Without Regeneration:Process 1-2s is isentropic i.e.,

But

Process 3-4s is isentropic

But

rr

s

PP

TT

1

1

2

1

2

KT s0

4.14.0

2 14.4284288

KTT

eiTTTT s

C0

2212

12 87.452288

28814.42885.0.,.

96.55441825.,.

4.14.0

4

1

3

4

3

4

s

rr

s TPP

TT

ei

KTT

eiTTTT

st

04

4

43

43 57.60396.554825

82582.0.,.

(i) Compressor work, WC = CP (T2 – T1)

= 1.005 (452.87 – 288) = 165.69 kJ/kg (ii) Turbine work, Wt = CP (T3 – T4)

= 1.005 (825 – 603.57) = 222.54 kJ/kg (iii) Work ratio = = 0.255 

(iv) Thermal Efficiency , = 15.2%

Case2: With Regeneration:We have effectiveness,

T5 = 565.890K Heat supplied, Q H1 = Q5-3 = CP(T3 – T5)

= 1.005 (825 – 565.89)= 260.4 kJ/kg

= 0.218

Improvement in th due to regenerator = 0.436

i.e., 43.6%

87.45257.60387.452

75.0.,. 5

24

25

T

eiTTTT

4.26085.56

1

H

CTth Q

WW

152.0152.0218.0

2.The maximum and minimum pressure and temperatures of a gas turbine are 5 bar, 1.2 bar and 1000K and 300K respectively. Assuming compression and expansion processes as isentropic, determine the th

(a) when an ideal regenerator is incorporated in the plant and (b) when the effectiveness of the above regenerator is 75%.

Solution: P2 = P3 = 5 bar P1 = P4 = 1.2 bar

T3 = 1000K T1 = 300K

Process 1-2s is isentropic i.e.,

rr

s

PP

TT

1

1

2

1

2

KT s04.1

4.0

2 21.4512.1

5300

Process 3-4s is isentropic i.e.,

rr

s

PP

TT

1

3

4

3

4

KT s04.1

4.0

4 88.66452.11000

Ideal regenerator: i.e., T5 = T4

Heat supplied = CP (T3 – T5)

= 1.005 [1000 – 664.88] = 336.79 kJ/kg Wnet = WT – WC = CP (T3 – T4) – CP (T2 – T1)

= 1.005 [1000 – 664.88 – 451.21 + 300] = 183.91

= 0.546 or 54.6%

79.33691.183

H

netth Q

W

Regenerator with = 0.75 i.e., i.e.,

Heat supplied, QH = CP (T3 – T5)= 1.005 (1000 – 611.46) =

390.48kJ/kg

= 0.471 or 47.1%

dropetemperaturidealdropetemperaturactual

TTTT

24

2575.0

KTT 0

55 46.611

21.45188.66421.451

75.0

48.39091.183

H

netth Q

W

3.Solve the above problem when the adiabatic efficiencies of the turbine and compressor are 90% and 85% respectively.

4. A gas turbine plant uses 500kg of air/min, which enters the compressor at 1 bar, 170C. The compressor delivery pressure is 4.4 bar. The products of combustion leaves the combustion chamber at 6500C and is then expanded in the turbine to 1 bar. Assuming isentropic efficiency of compressor to be 75% and that of the turbine to be 85%, calculate (i) mass of the fuel required /min, of the CV of fuel is 39000KJ/Kg. (ii)net power output (iii)Overall thermal efficiency of the plant. Assume CP=1.13KJ/Kg-K,=1.33 for both heating and expansion.

Solution: P1 = 1 bar T1 = 2900K P2 = 4.4 bar T3 =

9230KC = 0.75 t = 0.85 , WN = ? , th ?

Calorific Value = 39000 kJ/kgProcess 1-2s is isentropic compression i.e.,

sec/33.8min/500 kgkgma

?fm

C

T

PorVTVTorVPVP

1

122

1112211

KPP

TT s0

4.14.0

1

1

212 02.4434.4290

But

12

12

TTTT s

C

i.e., KTT

02

2

03.494290

29002.44375.0

Process 3-4s is isentropic expansion i.e.,

KTPP

TT

ss 033.1

32.0

4

1

3

4

3

4 18.6394.4

1923

But

sT TT

TT

43

43

i.e., KTT 0

44 76.68118.639923

92385.0

(i) ?fm

We have 03.49492313.139000500.,.

23

fPf

a

mei

TTCCV

mm

fm = 6.21kg/min

(ii) WN = ? Compressor work, WC = CP (T2 – T1)

= 1.005 (494.03 – 290)= 205.05 kJ/kg

Turbine work, WT = CP (T3 – T4)= 1.13 (923 – 681.76)= 272.6 kJ/kg

WN = WT – WC = 67.55 kJ/kg Net work output per minute =

= (500+6.21) (67.55) = 34194.49 kJ/minPower output = 569.91 kW 

Nfa Wmm

(iii) th = ?Heat supplied, QH = CP (T3 – T2)

= 1.33 (923 – 494.03)= 570.53 kJ/kg

= 0.118 or 11.8%

53.57055.67

H

Nth Q

W

5. A gas turbine cycle having 2 stage compression with intercooling in between stages and 2 stages of expansion with reheating in between the stages has an overall pressure ratio of 8.

The maximum cycle temperature is 14000K and the compressor inlet conditions are 1 bar and 270C. The compressors have s of 80% and turbines have s of 85%.

Assuming that the air is cooled back to its original temperature after the first stage compression and gas is reheated back to its original temperature after 1st stage of expansion, determine (i) the net work output

(ii) the cycle th. 

Solution: T5 = 14000K T1 = 3000K, P1= 1 bar

  C1= 0.8 = C2, t1 = t2 = 0.85 ,T3 = T1 ,T7 = T5For maximum work output,

88

5

1

4

8

7

6

5

3

4

1

2 PP

PP

PP

PP

PP

PP

barPPPPessureteIntermedia

83.2,Pr

7632

For process 1-2, 1

1

212

PP

TT s= 300 (2.83)0.286 = 403.950K

But KTTTT

TT sc

02

212

121 9.429

30030095.4038.0

Since T3 = T1 and 1

2

3

4

PP

PP

We have T4s = T2s = 403.950K

Also since C1 = C2, T4 = T2 = 429.90K

Compressor work, WC = CP (T2 – T1) + CP (T4 – T3) = 2 CP (T2 – T1) = 2 (1.005) (429.9 – 300) = 261.19 kJ/kg

For process 5 – 6,

KTPP

TT

ss 0

286.0

6

1

5

6

5

6 72.103983.211400

But KT

Tei

TTTT

st

06

6

65

651 76.1093

72.103914001400

85.0.,.

Since T7 = T5 and

8

7

6

5

PP

PP

, then T8 = T6

Since t1 = t2,T6 = T8 = 1093.760K

Turbine work, Wt = CP (T5 – T6) + CP (T7 – T8)= 2 CP (T5 – T6)= 2 (1.005) (1400 – 1093.76)= 615.54 kJ/kg

WN = WT – WC = 354.35 kJ/kg

th = ? Heat Supplied,

QH = CP (T5 – T4) + CP (T7 – T6)

= 1.005 (1400 – 429.9 + 1400 – 1093.76)

= 1282.72 kJ/kg

72.128235.354

th = 0.276 or 27.6%

6. Determine the of a gas turbine having two stages of compression with intercooling and two stages of expansion with reheat. Given that the pressure ratio is 4, minimum temperature of the cycle 270C and maximum temperature of the cycle is 6000C, when t, C and regenerator are equal to 80%.

( Home work) 7. A two stage gas turbine cycle receives air at 100 kPa

and 150C. The lower stage has a pressure ratio of 3, while that for the upper stage is 4 for the compressor as well as the turbine. The temperature rise of the air compressed in the lower stage is reduced by 80% by intercooling. Also, a regenerator of 78% effectiveness is used. The upper temperature limit of the cycle is 11000C. The turbine and the compressor s are 86%. Calculate the mass flow rate required to produce 6000kW.

Solution: P1 = 1 bar T1 = 2880K

IC = 0.8 ε = reg = 0.78, T5 = 13730K, C1 = C2 = t1 = t2 = 0.86, if P = 6000 kW

4,33

4

1

2 PP

PP

?m

Process 1-2s is isentropic compression

1

1

2

1

2

PP

TT s T2s = 288 (3)0.286

= 410.750K

But KTT

eiTTTT s

C0

2212

121 73.430

28828875.41086.0.,.

Also, KT

Tei

TTTT

IC0

33

12

32 54.31628873.430

73.4308.0.,.

Process 3-4s is 2nd stage isentropic compression

T4s = 316.54 (4)0.286 = 470.570K 

1

3

4

3

4

PP

TT s

But KTT

eiTTTT s

C0

4434

342 64.495

54.31654.31657.47086.0.,.

1

5

6

5

6

PP

TT s

KT s0

286.0

6 59.923411373

Process 5-6s is 1st stage isentropic expansion

But  

Process 6-7 is reheating, assume T7 = T5 = 13730K

KTT

eiTTTT

st

06

6

65

651 51.986

59.92313731373

86.0.,.

1

7

8

7

8

PP

TT s

KT s0

286.0

8 79.1002311373

Process 7-8s is 2nd stage isentropic expansion i.e.,

But KTT

eiTTTT

st

08

8

87

872 63.1054

79.100213731373

86.0.,.

Regenerator is used to utilizes the temperature of exhaust gases i.e.,

48

4

TTTTx

64.49563.105464.495

78.0.,.

xTei Tx = 931.650K

We have, Compressor work: WC = CP (T2 – T1) + CP (T4 – T3)= 1.005 (430.73 – 288 + 495.64 – 316.54)

= 323.44 kJ/kgAlso, Turbine work : WT = CP (T5 – T6) + CP (T7 – T8)

= 1.005 (1373 – 986.51 + 1373 – 1054.63)

= 708.38 kJ/kgNet work output, WN = WT - WC

= 384.95 kJ/kgBut, power produced,

i.e., 6000 x 1000 = 384.95 x 1000 = 15.59 kg/sec

We have, heat supplied, QH = CP (T5 – Tx) + CP (T7 – T6)= 1.005 (1373 – 931.65 + 1373 – 986.51)= 831.98 kJ/kg

NWmP

m

%3.46463.0 orQW

H

Nth

8. In a gas turbine plant working on Brayton cycle, the inlet conditions are 1 bar and 270C. The compression of air is carried out in two stages with a pressure ratio of 2.5 for each stage with intercooling to 270C.

The expansion is carried out in one stage with a pressure ratio of 6.25.

The maximum temperature in the cycle is 8000C. The of turbine and both compression stages are 80%. Determine (i) compressor work, (ii) Turbine work, (iii) Heat supplied, (iv) cycle , (v) cycle air rate.

  Hint: P1 = 1 bar P4 = P5 = 6.25 bar, P3 = P2 = 2.5 bar

9. The pressure ratio of an open cycle constant pressure gas turbine is 6. The temperature range of the plant is 150C and 8000C. Calculate (i) th of the plant, (ii) Power developed by the plant for an air circulation of 5 kg/s, (iii) Air fuel ratio, (iv) specific fuel consumption. Neglect losses in the system. Use the following data: for both air and gases: CP 1.005 kJ/kg0K and = 1.4. Calorific value of the fuel is 42000 kJ/kg, C = 0.85, t = 0.9 and combustion of 0.95.

10. In a G.T. unit with two stage compression and two stage expansion the gas temperature at entry to both the turbines are same. The compressors have an intercooler with an effectiveness of 83%. The working temperature limits are 250C and 10000C, while the pressure limits are 1.02 bar and 7 bar respectively. Assuming that the compression and expansion processes in the compressors and turbine are adiabatic with C of 84% and t of 89% for both the stages. Calculate (i) the air-fuel ratio at the combustion chambers if the calorific value of the fuel is 38500 kJ/kg, (ii) Power output in kW for an air flow rate of 1kg/s and (iii) overall cycle .

11. In a reheat gas turbine cycle, comprising one compressor and two turbine, air is compressed from 1 bar, 270C to 6 bar. The highest temperature in the cycle is 9000C. The expansion in the 1st stage turbine is such that the work from it just equals the work required by the compressor. Air is reheated between the two stages of expansion to 8500C. Assume that the isentropic s of the compressor, the 1st stage and the 2nd stage turbines are 85% each and that the working fluid is air and calculate the cycle .

Solution: P1 = 1 bar T1 = 300K P2 = 6 bar

T3 = 1173K WT1 = WC T5 = 1123K C = 0.85t1 = t2 = 0.85

We have process 1-2 is isentropic i.e.,

1

1

2

1

2

PP

TT S

KT S 5.50016300

4.14.0

2

KTT

eiTTTT

But SC 536

3003005.50085.0.,. 2

212

12

Compressor work, WC = CP (T2 – T1) = 1.005 (536 – 300) = 237 kJ/kg

From data, WT1 = WC = 237 kJ/kg = CP (T3 – T4) T4 = 937 kJ/kg

KTT

eiTTTT

But SSS

t 8951173

937117385.0.,. 4443

431

Process 3-4 is isentropic i.e., 1

3

4

3

4

TT

PP S

barP 328.211738956

4.04.1

4

From T-S diagram, intermediate pressure, P4 = P5 = 2.328 bar

Process 5-6s is isentropic in the 2nd stage turbine

KTPP

TT

ei SS 882

328.211123.,.

4.14.0

6

1

5

6

5

6

KTTeiTTTTButS

t 9188821123

112385.0.,. 66

65

652

WT2 = CP (T5 – T6)

= 1.005 (1123 – 918) = 206 kJ/kgNet work output = WT – WC

= (WT1 + WT2) – WC = 206 kJ/kgNet heat transfer or heat supplied, Q = QH + QR

= CP (T3 – T2) + CP (T5 – T4)= 640 + 187 = 827 kJ/kg

Cycle efficiency, %25

827206

net

netcycle Q

W

12. In a simple gas turbine unit, the isentropic discharge temperature of air flowing out of compressor is 1950C, while the actual discharge temperature is 2400C. Conditions of air at the beginning of compression are 1 bar and 170C. If the air-fuel ratio is 75 and net power output from the unit is 650kW. Compute (i) isentropic of the compressor and the turbine and (ii) overall . Calorific value of the fuel used is 46110 kJ/kg and the unit consumes 312 kg/hr of fuel. Assume for gases CP = 1.09 kJ/kg-K and = 1.32 and for air CP = 1.005 kJ/kg-K and = 1.4.

Solution: T2S = 195+273 = 468 K T2 = 240+273 = 513K

T1 = 290K P1=1bar A/F = 75, Power output = Wnet = WT – WC = 650kW C = ? T = ? cycle = ? CV = 46110 kJ/kg, CPg = 1.09 kJ/kg-k, g = 1.30, CPa = 1.005 kJ/kg-K, a = 1.4

skghrkgm f /0867.0/312

We have, Compressor Efficiency,

79.0290513290468.,.

12

12

eiTTTT S

C

Also, fa mFAm

= 75 (0.0867) = 6.503 kg/s

34.5290468Pr

4.04.1

1

1

2

TTRratioessure S

Applying SFEE to the constant pressure heating process 2-3,

23 TTCmmCVm Pgfaf

0.0867 (46110) = (6.503 + 0.0867) 1.09 (T3 – 513)T3 = 1069.6K

Also,

32.1132.1

4

1

3

4

3

4 34.56.1069

S

S TPP

TT g

g

T4S = 712.6K. Further,

1243 TTCmTTCmmWWW PaaPgfaCTnet

i.e., 650 = (6.503 + 0.0867) 1.09 (1069.6 – T4) – 6.503 (1.005) (513 – 290)T4 = 776K

822.06.7126.1069

7766.1069

43

43

S

T TTTT

163.0461100867.0

650

CVmW

f

netcycle

162.09.3997

650650

23

TTCmmQ

W

PgfaH

netcycle

Now, Turbine Efficiency,

And,

Or